If p^x = q^y = r^z , where x , y and z are in GP, then consider the...

Let $p^x = q^y = r^z = k$

Taking natural logarithm on all sides:

$x \ln p = y \ln q = z \ln r = \ln k$

$x = \dfrac{\ln k}{\ln p}$, $y = \dfrac{\ln k}{\ln q}$, $z = \dfrac{\ln k}{\ln r}$

Since $x$, $y$, and $z$ are in GP, $y^2 = xz$

Substituting the values of $x$, $y$, and $z$:

\left( \dfrac{\ln k}{\ln q} \right)^2 = \left( \dfrac{\ln k}{\ln p} \right)\left( \dfrac{\ln k}{\ln r} \right)$<br><br>$\dfrac{1}{(\ln q)^2} = \dfrac{1}{\ln p \ln r}$<br><br>$(\ln q)^2 = \ln p \ln r$<br><br>This implies that$\ln p$,$\ln q$, and$\ln r$ are in GP.

Statement II is correct, while Statement I is not necessarily true.

Answer: II only